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Friday, 24 February 2017

Law of Motion XIII


Physical Application
Horizontal Plane
i) Body kept on horizontal plane is at rest and no force is applied.
For vertical equilibrium
N = mg
ffriction = 0 (friction is a opposing force and there is no external applied force)
ii) Body kept on horizontal plane is at rest under single horizontal force.
For vertical equilibrium
N = mg (since body is at rest)
For horizontal equilibrium (since body is at rest)
F = fs
iii) Body kept on horizontal plane is just about to move.
For vertical direction N
N = mg (since body is at rest)
For horizontal direction (since body is just about to move) fs = fs(max) = µsN
F = fs = fs(max) = µsN
iv) Body kept on horizontal plane is accelerating horizontally.
For vertical direction N
N = mg (since body is at rest)
For horizontal direction
F – fk = ma
or, F – µkN = ma
v) Body kept on horizontal plane is accelerating horizontally towards right under single upward inclined force.
For vertical direction
N + FSinθ = mg (since body is at rest)
For horizontal direction FCosθ
FCosθ - fk = ma
or, FCosθ - µkN = ma
vi) Body kept on horizontal plane is accelerating horizontally towards right under single downward inclined force.
For vertical direction
N + FSinθ = mg (since body is at rest)
For horizontal direction
FCosθ - f= ma
or, FCosθ - µkN = ma
vii) Body kept on horizontal plane is accelerating horizontally towards right under an inclined force and opposing horizontally applied force
For vertical direction
N + FSinθ = mg (since body is at rest)
For horizontal direction
FCosθ - F1 - fk = ma
or, FCosθ - F- µkN = ma
viii) Body kept on horizontal plane is accelerating horizontally towards right under two inclined forces acting on opposite sides.
For vertical direction(since body is at rest)
N +  F1Sinθ = mg + F2 SinФ
For horizontal direction
F1Cosθ – F2CosФ - µkN = ma
Inclined Plane
i) Case - 1 
Body is at rest on inclined plane.
Perpendicular to the plane
N = mgCosθ (since body is at rest)
Parallel to the plane (since body is at rest)
mgSinθ = fs

ii) Case - 2
Body is just about to move on inclined plane. N
Perpendicular to the plane
N = mgCosθ (since body is at rest)
Parallel to the plane (since body is at rest)
mgSinθ = fs = fs(max) = µsN
iii) Case - 3
Body is accelerating downwards on inclined plane. N
Perpendicular to the plane
N = mgCosθ (since body is at rest)
Parallel to the plane
mgSinθ - fk = ma
or, mgSinθ - µkN = ma
iv) Case - 4
Body is accelerating up the incline under the effect of force acting parallel to the incline.
Perpendicular to the plane
N = mgCosθ (since body is at rest)
Parallel to the plane mgSinθ θ
F - fk - mgSinθ = ma
or, F - µk- mgSinθ = ma
v) Case - 5
Body accelerating up the incline under the effect of horizontal force.
Perpendicular to the plane
N = mgCosθ + FSinθ (since body is at rest)
Parallel to the plane
FCosθ - mgSinθ - fk = ma
or, FCosθ - mgSinθ - µkma

Vertical Plane
i) Case - 1
Body pushed against the vertical plane by horizontal force and is at rest.
For horizontal direction (since body is at rest)
F = N
For vertical direction N
mg = fs
ii) Case - 2
Body pushed against the vertical plane by horizontal force and pulled vertically upward
For horizontal direction (since body is at rest)
F = N F
For vertical direction
F1 - mg – fs = ma

iii) Case - 3
Body pushed against the vertical plane by inclined force and accelerates vertically upward.
For horizontal direction θ
N = FSinθ (since body is at rest)
For vertical direction
FCosθ - mg - fs = ma

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